MHB Guarantee positiveness of an expression

[Siron]

Hello!

I have the following expression:
$$\delta^2 = \frac{-\mu^2(5a\mu-4\mu^2-b)}{15a\mu-20\mu^2-3b}.$$
where $\mu \in \mathbb{R}$. Since $\delta^2$ is real (in the subject I'm working on) I need to check under which condition(s) the expression is well defined, that is, the RHS is positive and the denominator is not zero. The latter is not important for my question. Under certain assumption(s) on $a$ and $b$ it is not that difficult to find an interval around $\mu$ such that the RHS is indeed positive. However, it would be more suitable if I have conditions on $a$ and $b$ only (thus independent of $\mu$) such that the positiveness is guaranteed. I'm wondering if this is possible ...

In my opinion the big issue is the $-\mu^2$ in the numerator since this implies that I have to check:
$$\frac{5a\mu-4\mu^2-b}{15a\mu-20\mu^2-3b}<0$$
The expression is negative if and only if the numerator is positive and the denominator is negative (the other case is not possible here). Both numerator and denominator represent upside down parabola. If I write the denominator in vertex form then I guess I could force conditions on $a$ and $b$ such that it has a negative vertex and a negative discriminant and thus always negative values. However since the numerator is also an upside down parabola it is impossible to guarantee that it is always positive.

Any ideas?
Thanks in advance!

 

[Ackbach]

Hello!

I have the following expression:
$$\delta^2 = \frac{-\mu^2(5a\mu-4\mu^2-b)}{15a\mu-20\mu^2-3b}.$$
where $\mu \in \mathbb{R}$. Since $\delta^2$ is real (in the subject I'm working on) I need to check under which condition(s) the expression is well defined, that is, the RHS is positive and the denominator is not zero. The latter is not important for my question. Under certain assumption(s) on $a$ and $b$ it is not that difficult to find an interval around $\mu$ such that the RHS is indeed positive. However, it would be more suitable if I have conditions on $a$ and $b$ only (thus independent of $\mu$) such that the positiveness is guaranteed. I'm wondering if this is possible ...

In my opinion the big issue is the $-\mu^2$ in the numerator since this implies that I have to check:
$$\frac{5a\mu-4\mu^2-b}{15a\mu-20\mu^2-3b}<0$$
The expression is negative if and only if the numerator is positive and the denominator is negative (the other case is not possible here). Both numerator and denominator represent upside down parabola. If I write the denominator in vertex form then I guess I could force conditions on $a$ and $b$ such that it has a negative vertex and a negative discriminant and thus always negative values. However since the numerator is also an upside down parabola it is impossible to guarantee that it is always positive.

Any ideas?
Thanks in advance!

I think your plan in the last paragraph is the best you can do. As you say, you simply can't make the numerator of your fraction positive everywhere. However, I do see a problem even with the plan in your last paragraph: if you clamp down what $a$ and $b$ are by forcing the discriminant of the denominator to be negative, as well as the denominator having a negative vertex, that may not give you much flexibility in controlling how the numerator behaves, since the same $a$ and $b$ show up there. Theoretically, without considering the denominator for a second, you can control the shifting of the parabola in the numerator: $a$ controls the horizontal, and $b$ the vertical. Therefore, for any contiguous $\mu$ interval you desire, you can make the numerator positive on that interval. Which brings up the important point: can you restrict your attention to $\mu$ only in one interval? Suppose you have an interval $(x,y)$ to which you can restrict your $\mu$. You might possibly be able to find a non-empty intersection of the two sets:
$$\{(a,b):[(225a^2-4(20)(3b)<0] \, \land \, [(3b/20-9a^2/64)>0]\}$$
and
$$\{(a,b):-4\mu^2+5a\mu-b>0 \; \forall \, \mu\in(x,y)\}.$$

 

[Siron]

I think your plan in the last paragraph is the best you can do. As you say, you simply can't make the numerator of your fraction positive everywhere. However, I do see a problem even with the plan in your last paragraph: if you clamp down what $a$ and $b$ are by forcing the discriminant of the denominator to be negative, as well as the denominator having a negative vertex, that may not give you much flexibility in controlling how the numerator behaves, since the same $a$ and $b$ show up there. Theoretically, without considering the denominator for a second, you can control the shifting of the parabola in the numerator: $a$ controls the horizontal, and $b$ the vertical. Therefore, for any contiguous $\mu$ interval you desire, you can make the numerator positive on that interval. Which brings up the important point: can you restrict your attention to $\mu$ only in one interval? Suppose you have an interval $(x,y)$ to which you can restrict your $\mu$. You might possibly be able to find a non-empty intersection of the two sets:
$$\{(a,b):[(225a^2-4(20)(3b)<0] \, \land \, [(3b/20-9a^2/64)>0]\}$$
and
$$\{(a,b):-4\mu^2+5a\mu-b>0 \; \forall \, \mu\in(x,y)\}.$$

Thank you for your help Ackbach, I really appreciate it.

The two conditions of the first set are equivalent with $a^2 < \displaystyle \frac{16}{15}b$, which excludes $b<0$. Solving the inequality in the second set yields the following solution:
$$\mu \in \left]\frac{1}{8}\left(5a-\sqrt{25a^2-16b}\right), \frac{1}{8}\left(5a+\sqrt{25a^2-16b}\right)\right[$$
where $a \in \mathbb{R}$ and $b<\displaystyle \frac{25a^2}{16}$. The intersection of both sets is then given by:
$$\left \{a \in \mathbb{R}, b \in \left ] \frac{15}{16}a^2, \frac{25}{16}a^2\right[ \right \}$$
under the assumption that $\mu \in ]x,y[$ where $x$ and $y$ are given as above.

 

[Ackbach]

Thank you for your help Ackbach, I really appreciate it.

The two conditions of the first set are equivalent with $a^2 < \displaystyle \frac{16}{15}b$, which excludes $b<0$. Solving the inequality in the second set yields the following solution:
$$\mu \in \left]\frac{1}{8}\left(5a-\sqrt{25a^2-16b}\right), \frac{1}{8}\left(5a+\sqrt{25a^2-16b}\right)\right[$$
where $a \in \mathbb{R}$ and $b<\displaystyle \frac{25a^2}{16}$. The intersection of both sets is then given by:
$$\left \{a \in \mathbb{R}, b \in \left ] \frac{15}{16}a^2, \frac{25}{16}a^2\right[ \right \}$$
under the assumption that $\mu \in ]x,y[$ where $x$ and $y$ are given as above.

Yeah, I was just about to write that the two conditions in the first set are equivalent.

Your working looks correct to me! I think that's the best you can do.